I'm trying to better understand Ohm's law as it relates to general microelectronics projects by taking a simple project I'm working on now.
I have a 9V battery powering a small tower pro sg90 servo. The specs on the motor is that it takes 5Volts, and has a running current draw of about 220 milliamps (+-50). So if I use ohm's law to calculate the resistance needed if I'm supplying 9V, that'd be 9V=.220*x or 40 ohms. That seems really small to add a resistor on. The smallest I can find are about 110ohm resistors.
Should I add a resistor before the motor? And if so, how do I calculate what size I need?
Best Answer
Assuming you should do this, your math is a little off.
\$R=V_{Drop}/I_{Min} = 4/(.220-0.05) = 23.5\Omega\$
Power \$P = I_{Max}^2R = (.220+0.05)^2*23.5 = 1.7W\$
Be aware though...
Start current will be much higher than that, with the above R in place closer to \$9/23.5 = .380A\$. That means a stalled motor will cause ALMOST 3.5W of heat to be dissipated in the resistor. So you need to up the wattage to closer to 4W.
And yes those resistors are quite small. You may need to put several in parallel to obtain the value and power rating you need. Ultimately, using a resistor for this purpose is not the best way to go.
The biggest issue you will have with it is your startup current will be limited to the value I mentioned. That means the acceleration will be severely limited and in some cases, the motor may not even start.
As such you really ought to be powering the motor with something more active. A simple 1A linear voltage regulator would work well, provided you have enough heat-sinking to keep it cool. Or you could use a simple voltage follower circuit like the one below.
simulate this circuit – Schematic created using CircuitLab
Ultimately though a buck-boost regulator would get you the most efficiency and power from your battery.